Weird problem!

[@reaperskyfall6691]

That one substitution changed whole game

[@owlsmath]

yep! :) 👍👍👍

[@Dharun-ge2fo]

You could have directly differentiated u and then solve for root(x+1) and root (x) by adding and subtracting the equations, u= root(x+1)-root(x) , and (1/u)= root(x+1)+root(x)

[@LipschitzHutchinson]

An alternative method would be to substitute x=sinh^2 u, although it's a bit messier since you need to do some simplifications with arcsinh.
Although, now that I think about it, it really is just equivalent to setting x=(u-1/u)^2/4, which is what was done in the problem.

[@owlsmath]

Makes sense. Thanks 🙏

[@dkravitz78]

Just to simplify a little. Let v be the '+' version of u.
uv=1 and v-u=2 sqrt (x)
so 4x = v^2-2uv+u^2 = 1/u^2 -2 + u^2
From there 4 dx is easy to see -1/u^3+2u times du
I don't know if I'd call that a huge game but it does simplify a little

[@owlsmath]

Thats nice! Its easier to differentiate that way but I also like that you get to use that relationship between u and v that i mentioned briefly at the beginning of the video.

[@slavinojunepri7648]

Fantastic

[@owlsmath]

Thanks! 👍

[@maxvangulik1988]

x=sinh^2(t)
sqrt(x+1)=cosh(t)
I=int(e^-(pi•t))dt
I=-1/pi•e^-(pi•t)+C
t=arsinh(sqrt(x))=ln(sqrt(x)+sqrt(x+1))
I=-1/pi•(sqrt(x+1)-sqrt(x))^pi+C

[@owlsmath]

thanks! Nice method :)

[@MikeMagTech]

Nice job. It simplified nicely, but getting there was not straightforward.

[@owlsmath]

Thanks! Yep it’s a tricky substitution in this one. 👍